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Question

In figure, the coefficient of static friction between the floor and block B is 0.30. The coefficient of static friction between the block B and A is 0.15. The mass of block A is m2 kg and B is m kg. What is the maximum horizontal force F in Newtons that can be applied to block B, so that the two blocks move together?



A
1.35 mg
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B
0.45 mg
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C
0.675 mg
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D
0.9 mg
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Solution

The correct option is C 0.675 mg
mA=m/2 kg,  mB=m kg
μA=0.15,    μB=0.3
Let both the blocks are moving together with common acceleration a. The maximum force will correspond to the maximum value of f1, which can sustain the required acceleration for block A


From FBD of block,
maximum value of f1=μANA=μA=μamAg
 
a=μAmAgmA=μAg=0.15g m/s2 ...(i)
Now, on the system of blocks A+B, applying equation of dynamics from the FBD:


Maximum value of f2=μBN=μB(mA+mB)g
Hence applying newton's 2nd law,   
FμB(mB+mA)g=(mB+mA)a
putting a=0.15g from Eq. (i)
F=0.3(mB+mA)g+(mB+mA)0.15g
F=0.45(mB+mA)g 
F=(m+m2)0.45g
F=32m×0.45g=0.675 mg N

Physics

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